My question for today is-
Consider the function $f : \mathbb{R}^3 → \mathbb{R}$ defined by:
$f(x,y,z)=\left\{ \begin{array}{c l} \frac{z}{x}+y &if &x≠0 \\ a & if &x=0 \\ \end{array}\right.$
where $a ∈ \mathbb{R}$ is some fixed constant. Determine whether there exists a value of $a ∈ \mathbb{R}$ such that $f(x, y, z)$ is continuous at the origin.
This question is one im not really sure where to go to obtain an $a$ value so any help will be appreciated.
No. Suppose for contradiction that there exists $a \in \mathbb{R}$ such that $f$ is continuous at $0$. Then we have
$$a = f(0,0,0) = \lim_{x \to 0} f(x,0,0) = \lim_{x \to 0} \left(\frac{0}{x} + 0\right) = 0$$
and
$$a = f(0,0,0) = \lim_{x \to 0} f(x,x,x) = \lim_{x \to 0} \left(\frac{x}{x} + x\right) = \lim_{x \to 0} (1+x) = 1,$$
so $0=1$, a contradiction.
Fix any real $a$ and define $f$ piecewise with this $a$, as you have written above.
For the sake of contradiction, suppose $f$ is continuous at the origin. This means that for any sequence of points $\mathbf{x}_n \in \mathbb{R}^3$ converging to the origin, we must have that $\lim_{n \to \infty} f(\mathbf{x}_n)$ exists and equals $f(\mathbf{0}) = a$.
What then about the sequence defined by $\mathbf{x}_n = \left(\frac{1}{n^2},0,\frac{1}{n} \right)$? For this sequence, $f(\mathbf{x}_n) = \frac{\frac{1}{n}}{\frac{1}{n^2}} + 0 = n$, so $\lim_{n \to \infty}f(\mathbf{x}_n)$ does not exist.
By contradiction, it cannot be that $f$ is continuous at the origin.